Question

According to data from the state blood program, 40 percent of all individuals have group A blood. If six individuals give blood, find the probability that exactly three of the individuals have group A blood.

Answer 1

Answer: 0.27648

Step-by-step explanation:

Given : The proportion of all  individuals have group A blood : p=0.040

Total individuals give blood : n= 6

Let X be the number of individuals have group A blood.

Since all individual are independent of each other.

$X\sim Bin(n=6, p=0.40)$

Formula : $P(X=x)=^nC_xp^x(1-p)^{n-x}$ , where n= sample size , p = probability of getting success in each trial.

The probability that exactly three of the individuals have group A blood. :

$P(x=3)= ^6C_3(0.40)^3(1-0.4)^3\\\\= \dfrac{6!}{3!3!}\times(0.40)^3(0.60)^3\\\\=0.27648$

The probability that exactly three of the individuals have group A blood. is 0.27648